The camera has pixel size of p=0.012 mm. What linear distance does this correspo

Question

The camera has pixel size of p=0.012 mm. What linear distance does this correspond to on the lunar surface? What angle does the pixel subtend?

7. (40 points) Assume you have a camera with an effective focal length of f-2500mm and that the detector is dx d-50mm x 50mm. The camera is on a spacecraft in circular orbit around the Earth's Moon at an altitude of h-25km a) (10 points) What are the dimensions (in length units) of the lunar surface area seen by the detector'? b) (10 points) The camera has pixel size of p-0.012 mm. What linear distance does this correspond to on the lunar surface? What angle does the pixel subtend? Give your answer in radians. Also, convert the radians to degrees. Convert the degrees to arcminutes. Convert the arcminutes to arcseconds, c) (10 points) If the camera looks straight down (at nadir), how long does it take for an object at one edge of the field-of-view (FOV) to move to the opposite edge? Assume the detector is mounted orthogonal to the direction of motion (i.e. the object doesn't move across the FOV at an angle)

Explanation / Answer

7. given camera effetive focallength

f = 2500 mm

d x d = 50 mm * 50 mm ( detector)

h = 25 km

Radius of moon, Rm = 1737,000 m

a. let aperture be D

then

f/D = (f + (Rm + h))/S

then

S = 704801D

D = 50 mm * 50 mm

S = 35240.05 m * 35240.05 m

b. p = 0.012 mm

on lunar surface this corresponds to

pl = 0.012*704801 = 8.457612 m

angle = theta = pl/(h+ Rm + f) = 0.0000048 rad = 2.75*10^-4 deg = 0.016501 min = 0.990071 sec

c. we need nadir's speed to calculate time taken for him to cover d = 35240.05 m

let his speed be v

then time taken = t

t = d/v = 35240.05/v s

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